Let $(\Omega,\mathcal A,\mu)$ be a measure space, $E$ be a normed space and $f:\Omega\to E$ be strongly$^1$ $\mathcal A$-measurable with $$\langle f,\varphi\rangle\in\mathcal L^1(\mu)\;\;\;\text{for all }\varphi\in E'.$$
Are we able to show that $$\int\langle f,\varphi\rangle\:{\rm d}\mu=\langle x,\varphi\rangle\;\;\;\text{for all }\varphi\in E'\tag1$$ for some $x\in E$? If not, what are sufficient conditions to ensure that?
If $f$ has finite range, the claim is obviously true. If not, there are $\mathcal A$-measurable $f_n:\Omega\to\mathbb R$ with finite range for $n\in\mathbb N$ satisfying $\left\|f_n\right\|_E\le\left\|f\right\|_E$ and $\left\|f_n-f\right\|_E\xrightarrow{n\to\infty}0$. So, there are $(x_n)_{n\in\mathbb N}\subseteq E$ with $$\int\langle f_n,\varphi\rangle\:{\rm d}\mu=\langle x_n,\varphi\rangle\;\;\;\text{for all }\varphi\in E'\text{ and }n\in\mathbb N\tag2.$$ Using Lebesgue’s dominated convergence theorem, it's now easy to see that $$\langle x_n,\varphi\rangle\xrightarrow{n\to\infty}\int\langle f,\varphi\rangle\:{\rm d}\mu\tag3\;\;\;\text{for all }\varphi\in E',$$ which at least implies that $(x_n)_{n\in\mathbb N}$ is weakly Cauchy.
However, we know that a weak Cauchy sequence doesn't need to weakly converge. So, maybe the problem I'm asking here is a particular instance of the fact that if $(x_n)_{n\in\mathbb N}$ is any weakl Cauchy sequence, then $$L\varphi:=\lim_{n\to\infty}\langle x_n,\varphi\rangle\;\;\;\text{for all }\varphi\in E'$$ is a well-defined linear functional (since the field over which $E$ is a vector space is complete), but $L$ is not necessarily of the form $$L\varphi=\langle x,\varphi\rangle\tag4$$ for some $x\in E$.
Maybe all of this boils down to the properties of the canonical embedding $$\iota:E\to E''\;,\;\;\;x\mapsto\left(E'\ni\varphi\mapsto\langle x,\varphi\rangle\right).$$
In any case, I would really appreciate if someone could show under which conditions we are able to prove the desired result; but I'm also interested in results related to the general case for the functional $L$.
EDIT
As David Mitra pointed out in the comments, it seems like the question boils down to find conditions under which a Dunford integrable function is even Pettis integrable. Let me fix the terminology:
$f:\Omega\to E$ to is called Dunford $\mu$-integrable if $\langle f,\varphi\rangle\mathcal L^1(\mu)$. In that case, $$\int f\:{\rm d}\mu:E'\to\mathbb K\;,\;\;\;\varphi\mapsto\int\langle f,\varphi\rangle\:{\rm d}\mu\tag5$$ is a bounded linear functional on $E'$, i.e. $$\int f\:{\rm d}\mu\in E''.\tag6$$ Now $f$ is called Pettis $\mu$-integrable if $f$ is Dunford $\mu$-integrable and $$\int f\:{\rm d}\mu\in\iota E.\tag7$$
First approach for a positive result: I've often seen the assertion that the desired claim is true when $\Omega=[a,b]$ for some $a<b$, $\mathcal A=\mathcal B([a,b])$, $\mu$ is the restriction of the Lebesgue measure on $\mathcal B(\mathbb R)$ to $[a,b]$ and $f$ is continuous. The crucial argument is that $[a,b]$ is compact, hence $f$ is even uniformly continuous, which allows us to find $(x_n)_{n\in\mathbb N}\subseteq E$ such that $$\langle x_n,\varphi\rangle\xrightarrow{n\to\infty}\int\langle f,\varphi\rangle\:{\rm d}\mu\tag8$$ uniformly with respect to $\left\|\varphi\right\|_{E'}\le1$. By Lemma 1 below, this is enough to conclude, but I'm quite sure that this result holds in way more generality.
Lemma 1: In general, $(x_n)_{n\in\mathbb N}\subseteq E$ is norm convergent if and only if $$\sup_{\left\|\varphi\right\|_{E'}\le1}\left|\langle x_m-x_n,\varphi\rangle\right|\xrightarrow{m,\:n\to\infty}0.$$
Second approach for a positive result: Let me continue with the setting I've described at the beginning of this post, i.e. the situation where $f$ is strongly $\mathcal A$-measurable.
Most probably I'm missing something, but if $(x_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ are as in $(2)$ and $(3)$, then we should have \begin{equation}\begin{split}\sup_{\left\|\varphi\right\|_{E'}\le1}\left|\langle x_m-x_n,\varphi\rangle\right|&=\sup_{\left\|\varphi\right\|_{E'}\le1}\left|\int\langle f_m-f_n,\varphi\rangle\:{\rm d}\mu\right|\\&\le\int\left\|f_m-f_n\right\|_E\:{\rm d}\mu\xrightarrow{m,\:n\:\to\:\infty}0\end{split}\tag9.\end{equation} So, by Lemma 1, we should be able to conclude. Am I missing something?
If not and this is correct, is this somehow a generalization of what I've described in the "first approach"? Therein $f$ is continuous (hence $\mathcal A$-measurable) and, since $[a,b]$ is separable, $f([a,b])$ is separable. So, $f$ is strongly $\mathcal A$-measurable. Maybe that's the crucial point.
$^1$ i.e. $f$ is $\mathcal A$-measurable and $f(\Omega)$ is separable. The crucial point is that a function $f:\Omega\to E$ is strongly $\mathcal A$ if and only if there are $\mathcal A$-measurable $f_n:\Omega\to\mathbb R$ with finite range for $n\in\mathbb N$ satisfying $\left\|f_n\right\|_E\le\left\|f\right\|_E$ and $\left\|f_n-f\right\|_E\xrightarrow{n\to\infty}0$.